Exercise set

Vacuum and Rarefied Gas Systems Exercises

Worked vacuum exercises for pumpdown, gas load, RGA partial pressure, water exposure, purge dilution, hold time and guarded acceptance.

These exercises practise vacuum and rarefied gas engineering physics as calculation evidence. They connect pressure units, gas inventory, effective pumping speed, conductance, gas load, rate-of-rise testing, outgassing, residual-gas partial pressure, purge dilution, mean free path, Knudsen number, gauge placement and heat transfer in vacuum.

The goal is not only to reach a pressure number. The goal is to decide whether the measured vacuum state represents the region, gas species, surfaces, temperature and operating condition that matter for the device under test.

Assume simplified screening models unless an exercise states otherwise. Real vacuum work should also check pump curves, gas species correction, gauge calibration, valve state, conductance path, surface history, material compatibility, contamination risk, bakeout state, thermal gradients, leak-test sensitivity and operating procedures.

Release Evidence Notes

Use these exercises as screening evidence for vacuum acceptance, not as substitutes for a configured test record. A credible release should connect each pressure, gas-load, conductance, thermal or purge result to the chamber configuration, valve state, gauge location, gas species, surface history, temperature state and acceptance threshold.

The minimum evidence set is:

  • the pressure boundary being accepted: pump inlet, chamber body, process volume, local test article, isolated volume or trapped cavity;
  • gauge evidence with calibration gas, correction factor, range, location, conductance path, repeatability and uncertainty;
  • gas-load evidence that separates external leak, outgassing, virtual leak, process gas, purge gas and residual contamination where possible;
  • RGA, oxygen, moisture, rate-of-rise, pumpdown-curve or bakeout records when total pressure alone cannot own the release decision;
  • guard bands for calibration, repeatability, local-pressure inference, gas-species correction, partial-pressure limits, purge dilution and thermal-vacuum margins.

Treat a single pressure reading as weak evidence when location, gas composition or surface state controls the result. A pump-side gauge, a total-pressure pass, a nominal purge time or a room-temperature pumpdown can all hide the failure mode that matters at the test article.

How to Use These Exercises

For each exercise, define:

  1. the accepted pressure boundary: pump inlet, chamber body, local test article, process volume or isolated volume;
  2. whether pressure is absolute, gauge, total pressure or species-dependent;
  3. the gas species, residual-gas composition, temperature and relevant volume or conductance path;
  4. the acceptance threshold: pumpdown time, local pressure, gas load, leak-rate, outgassing, regime or thermal limit;
  5. the measurement evidence needed to trust the result.

The most common error is accepting a vacuum system from one gauge number. A credible review connects pressure, location, gas load, conductance, surfaces, temperature and instrument uncertainty.

Engineering Boundary Notes

These exercises use simplified vacuum and rarefied-gas screening models. They do not replace leak checking, calibrated gauge comparison, residual-gas analysis, pump curve review, bakeout validation, contamination-control review, thermal-vacuum testing or procedure qualification. A pressure result applies only to the stated volume, conductance path, gas species, surface state, temperature, valve configuration and gauge location.

Separate total pressure, partial pressure, local pressure and inferred process condition before release. A chamber-body gauge can pass while a restricted fixture, hot surface, trapped cavity or water-contaminated local region fails. Release evidence should state whether the accepted boundary is a pump inlet, chamber volume, local test article, process surface or isolated volume.

Common Release Mistakes

  • accepting a pump-side gauge reading as local test-article pressure without conductance evidence;
  • treating total pressure as contamination evidence when partial pressure or exposure dose controls the process;
  • using nominal pump speed while ignoring conductance, valve state, foreline limit and crossover interlocks;
  • calling a rate-of-rise result a leak rate before outgassing, virtual leaks and temperature changes are separated;
  • applying a gauge gas correction without recording gas species, range, calibration and uncertainty;
  • releasing after purge, pump trip or bakeout changes without updated operating-state evidence.

Scenario Map

ScenarioExercisesEngineering decision
Pressure state and inventory1Convert units and estimate gas inventory before interpreting a vacuum reading.
Pumping and conductance2, 3, 7, 13Separate pump nameplate speed from installed chamber speed, crossover pressure and local pressure.
Gas load and surface condition4, 5Distinguish leak, outgassing, fixture and rate-of-rise evidence.
Rarefied-flow regime6Check whether continuum assumptions are still credible for the local geometry.
Thermal vacuum behavior8, 17Decide whether radiation alone can remove heat and whether a room-temperature gauge represents a hot local volume.
Recovery and diagnosis9, 10Use pumpdown regression and finite trapped-volume inventory to avoid false leak conclusions.
Measurement acceptance11, 12, 14Apply gas-species correction, partial-pressure limits and uncertainty guard bands before releasing the system.
Surface contamination exposure18Convert water-vapor partial pressure and time into exposure dose
Purge and operating state15Check whether backfill or purge assumptions produce the required gas composition.
Operational upset and hold time16Convert gas load and pressure guard into pump-trip response time and interlock margin.

Validation Package Checklist

  • accepted pressure boundary, chamber configuration, valve state and conductance path are named;
  • gauge type, location, calibration gas, correction factor, range, repeatability and uncertainty are recorded;
  • leak, outgassing, virtual leak, process gas, purge gas and residual contamination evidence are separated;
  • pumpdown, rate-of-rise, RGA, moisture, oxygen, bakeout or purge records are included when total pressure is insufficient;
  • gas species, temperature, surface history, material exposure and local-pressure inference are bounded;
  • interlock, crossover, pump-trip hold time and operating-state assumptions are configuration controlled;
  • final release decision states accept, leak check, bake, purge, retest, restrict operation or hold.

Exercise 1: Pressure Units and Gas Inventory

A vacuum chamber has volume:

V=0.60\ \text{m}^3

The local high-vacuum gauge reads:

p=2.0\times 10^{-3}\ \text{Pa}

at:

T=293\ \text{K}

Convert the pressure to mbar and Torr. Estimate the number of moles and molecules in the chamber using an ideal-gas inventory.

Solution

Use:

1\ \text{mbar}=100\ \text{Pa}

Therefore:

\displaystyle p=\frac{2.0\times 10^{-3}}{100}=2.0\times 10^{-5}\ \text{mbar}

Use:

1\ \text{Torr}=133.322\ \text{Pa}

Therefore:

\displaystyle p=\frac{2.0\times 10^{-3}}{133.322}=1.50\times 10^{-5}\ \text{Torr}

Moles in the chamber are:

\displaystyle n=\frac{pV}{RT}
\displaystyle n=\frac{(2.0\times 10^{-3})(0.60)}{8.314(293)}=4.93\times 10^{-7}\ \text{mol}

Molecules in the chamber are:

\displaystyle N=\frac{pV}{k_BT}
\displaystyle N=\frac{(2.0\times 10^{-3})(0.60)}{(1.380649\times 10^{-23})(293)}=2.97\times 10^{17}

Engineering Comment

The pressure is very low compared with atmospheric pressure, but the chamber still contains many gas molecules. Vacuum engineering is not “empty space.” The result is useful for inventory reasoning, but it does not identify gas species, adsorption on surfaces, local pressure gradients, contamination risk or whether the gauge calibration gas matches the actual gas mixture.

Plausibility Check

The pressure conversion should reduce the Pa number by a factor of 100 when converting to mbar and by about 133 when converting to Torr. The gas inventory is less than a micromole, which is plausible for high vacuum, but the molecule count remains large enough that contamination and species composition still matter.

Exercise 2: Effective Pumping Speed Through a Conductance Restriction

A turbomolecular pump has rated nitrogen speed:

S_p=0.20\ \text{m}^3/\text{s}

The valve, baffle and connecting duct between the chamber and pump have molecular-flow conductance:

C=0.030\ \text{m}^3/\text{s}

Estimate the effective pumping speed at the chamber.

Solution

For a pump and conductance restriction in series:

\displaystyle \frac{1}{S_{eff}}=\frac{1}{S_p}+\frac{1}{C}

Substitute:

\displaystyle \frac{1}{S_{eff}}=\frac{1}{0.20}+\frac{1}{0.030}
\displaystyle \frac{1}{S_{eff}}=5.00+33.33=38.33\ \text{s/m}^3

Therefore:

S_{eff}=0.0261\ \text{m}^3/\text{s}

The effective speed is:

\displaystyle \frac{S_{eff}}{S_p}=\frac{0.0261}{0.20}=0.130

or about 13\% of pump nameplate speed.

Engineering Comment

The conductance path, not the pump nameplate, controls the chamber. This is why baffles, screens, long ducts, narrow valves and local enclosures can dominate pumpdown and local pressure. A vacuum acceptance report should state installed valve position and conductance assumptions, not only pump model number.

Plausibility Check

The effective speed must be lower than both the pump speed and the conductance because the restrictions act in series. A result near the conductance value is expected here because C=0.030\ \text{m}^3/\text{s} is much smaller than the pump nameplate speed.

Exercise 3: Ideal Pumpdown Time and Plateau Gas Load

Use the effective pumping speed from Exercise 2:

S_{eff}=0.0261\ \text{m}^3/\text{s}

A chamber volume is:

V=0.80\ \text{m}^3

If the chamber behaved as a simple well-mixed volume with no continuing gas load, estimate ideal pumpdown time from:

p_0=100\ \text{Pa}

to:

p_1=1.0\times 10^{-3}\ \text{Pa}

The measured pressure instead settles near:

p_{\infty}=3.0\times 10^{-3}\ \text{Pa}

Estimate the implied steady gas load at the plateau.

Solution

The ideal exponential pumpdown relation is:

\displaystyle p(t)=p_0\exp\left(-\frac{S_{eff}t}{V}\right)

Rearrange:

\displaystyle t=\frac{V}{S_{eff}}\ln\left(\frac{p_0}{p_1}\right)

Substitute:

\displaystyle t=\frac{0.80}{0.0261}\ln\left(\frac{100}{1.0\times 10^{-3}}\right)
t=30.7\ln(1.0\times 10^5)=353\ \text{s}
t=5.9\ \text{min}

The measured plateau indicates a continuing gas load. At steady state:

Q=p_{\infty}S_{eff}
Q=(3.0\times 10^{-3})(0.0261)=7.83\times 10^{-5}\ \text{Pa m}^3/\text{s}

Engineering Comment

The ideal time is much shorter than many real pumpdowns because it ignores desorption, water vapor, virtual leaks, conductance changes during roughing, gauge response and surface history. The plateau gas load is the more important diagnostic result. If the chamber stops improving at 3.0\times 10^{-3}\ \text{Pa}, a larger pump may not solve the problem unless the gas source and conductance path are understood.

Plausibility Check

The pressure ratio is 10^5, so several time constants are required even in an ideal model. The plateau load is larger than the Exercise 4 acceptance gas load, so the two results would trigger a consistency review if they came from the same hardware and surface state.

Exercise 4: Rate-of-Rise Gas Load and Acceptance

A chamber is isolated from the pump for a rate-of-rise test. The isolated volume is:

V=0.50\ \text{m}^3

Local pressure rises from:

p_1=1.2\times 10^{-3}\ \text{Pa}

to:

p_2=4.8\times 10^{-3}\ \text{Pa}

in:

\Delta t=900\ \text{s}

The acceptance limit for total gas load is:

Q_{max}=2.0\times 10^{-5}\ \text{Pa m}^3/\text{s}

A helium leak test estimates an external leak contribution of:

Q_{He}=4.0\times 10^{-7}\ \text{Pa m}^3/\text{s}

Estimate the total gas load and decide whether the system passes the gas-load limit. Also estimate what fraction of the measured gas load is explained by the external helium leak.

Solution

Pressure rise is:

\Delta p=p_2-p_1
\Delta p=4.8\times 10^{-3}-1.2\times 10^{-3}=3.6\times 10^{-3}\ \text{Pa}

The rate-of-rise gas load is:

\displaystyle Q=V\frac{\Delta p}{\Delta t}
\displaystyle Q=0.50\frac{3.6\times 10^{-3}}{900}=2.0\times 10^{-6}\ \text{Pa m}^3/\text{s}

Compare with the acceptance limit:

2.0\times 10^{-6}<2.0\times 10^{-5}

The total gas load passes the stated limit.

The helium leak fraction is:

\displaystyle f_{He}=\frac{Q_{He}}{Q}
\displaystyle f_{He}=\frac{4.0\times 10^{-7}}{2.0\times 10^{-6}}=0.20

The external helium leak explains about 20\% of the measured gas load.

Engineering Comment

Passing the total gas-load limit does not mean every source is acceptable. The helium contribution is small enough that outgassing or trapped volumes dominate the remaining 80\%, but the leak still matters if the process is sensitive to air, oxygen, water or helium background. Rate-of-rise tests should record valve state, temperature, gauge range, stabilization time and whether the chamber was recently vented.

Plausibility Check

The pressure rise is only 3.6\times 10^{-3}\ \text{Pa} over 15 minutes, so a small gas load is expected. The helium leak cannot explain the full rate-of-rise because its contribution is only one fifth of the measured total.

Exercise 5: Outgassing Area and Target Pressure

A cleaned chamber has internal surface area:

A=12\ \text{m}^2

The expected outgassing rate is:

q_A=2.5\times 10^{-7}\ \text{Pa m}^3/(\text{s m}^2)

An internal fixture contributes:

Q_{fixture}=6.0\times 10^{-6}\ \text{Pa m}^3/\text{s}

Use:

S_{eff}=0.0261\ \text{m}^3/\text{s}

Estimate total gas load and steady pressure. Decide whether the system can meet:

p_{target}=1.0\times 10^{-3}\ \text{Pa}

Solution

Surface outgassing load is:

Q_A=q_AA
Q_A=(2.5\times 10^{-7})(12)=3.0\times 10^{-6}\ \text{Pa m}^3/\text{s}

Total gas load is:

Q_{total}=Q_A+Q_{fixture}
Q_{total}=3.0\times 10^{-6}+6.0\times 10^{-6}=9.0\times 10^{-6}\ \text{Pa m}^3/\text{s}

Steady pressure is:

\displaystyle p_{\infty}=\frac{Q_{total}}{S_{eff}}
\displaystyle p_{\infty}=\frac{9.0\times 10^{-6}}{0.0261}=3.45\times 10^{-4}\ \text{Pa}

This is below:

1.0\times 10^{-3}\ \text{Pa}

so the system passes this simplified steady-pressure screen.

Engineering Comment

The margin is useful, but it depends strongly on surface condition. Fingerprints, absorbed water, polymer cables, adhesives, trapped volumes and warm fixtures can increase gas load by orders of magnitude. The calculation should be tied to cleaning evidence, bakeout history, material list and a rate-of-rise or residual-gas trend.

Plausibility Check

The fixture contributes twice the cleaned surface load, so fixture material and temperature are more important than the chamber wall area in this screen. The target is passed by about a factor of three, which is useful but not enough to ignore uncertainty in outgassing history.

Exercise 6: Mean Free Path and Knudsen Regime

Nitrogen in a chamber is at:

p=5.0\times 10^{-2}\ \text{Pa}
T=293\ \text{K}

Use molecular diameter:

d_m=3.7\times 10^{-10}\ \text{m}

Estimate mean free path:

\displaystyle \lambda=\frac{k_BT}{\sqrt{2}\pi d_m^2p}

Then compute Knudsen number for:

  1. a 10\ \text{mm} chamber feature;
  2. a 0.5\ \text{mm} sensor gap.

Solution

Mean free path is:

\displaystyle \lambda=\frac{(1.380649\times 10^{-23})(293)}{\sqrt{2}\pi(3.7\times 10^{-10})^2(5.0\times 10^{-2})}
\lambda=0.133\ \text{m}

For the 10\ \text{mm} feature:

L=0.010\ \text{m}
\displaystyle Kn=\frac{\lambda}{L}=\frac{0.133}{0.010}=13.3

For the 0.5\ \text{mm} gap:

L=0.0005\ \text{m}
\displaystyle Kn=\frac{0.133}{0.0005}=266

Engineering Comment

Both length scales are in the free molecular regime by this screen, and the sensor gap is much more strongly rarefied. This is why one chamber pressure can correspond to different local physics depending on geometry. Continuum correlations for convection or pressure drop would be inappropriate without regime correction or validation.

Plausibility Check

At 5.0\times 10^{-2}\ \text{Pa}, a mean free path on the order of centimeters to decimeters is plausible for nitrogen near room temperature. Both Knudsen numbers are much greater than 1, so any continuum-flow answer would be a red flag.

Exercise 7: Conductance-Limited Local Pressure

A chamber is connected to a pump through a long tube. In molecular flow for air near room temperature, approximate round-tube conductance in liters per second as:

\displaystyle C_{L/s}=12.1\frac{D_{cm}^3}{L_{cm}}

The tube has:

D=50\ \text{mm}
L=0.70\ \text{m}

The pump speed is:

S_p=0.18\ \text{m}^3/\text{s}

The gas load is:

Q=1.5\times 10^{-5}\ \text{Pa m}^3/\text{s}

Estimate tube conductance, effective chamber pumping speed, chamber pressure and pump-side pressure.

Solution

Convert dimensions:

D=5.0\ \text{cm}
L=70\ \text{cm}

Molecular conductance is:

\displaystyle C_{L/s}=12.1\frac{5.0^3}{70}=21.6\ \text{L/s}

Convert to cubic meters per second:

C=0.0216\ \text{m}^3/\text{s}

Effective chamber pumping speed is:

\displaystyle \frac{1}{S_{eff}}=\frac{1}{S_p}+\frac{1}{C}
\displaystyle \frac{1}{S_{eff}}=\frac{1}{0.18}+\frac{1}{0.0216}=5.56+46.3
S_{eff}=0.0193\ \text{m}^3/\text{s}

Chamber pressure is:

\displaystyle p_{ch}=\frac{Q}{S_{eff}}
\displaystyle p_{ch}=\frac{1.5\times 10^{-5}}{0.0193}=7.78\times 10^{-4}\ \text{Pa}

Pump-side pressure estimated from pump speed alone is:

\displaystyle p_{pump}=\frac{Q}{S_p}
\displaystyle p_{pump}=\frac{1.5\times 10^{-5}}{0.18}=8.33\times 10^{-5}\ \text{Pa}

The chamber pressure is:

\displaystyle \frac{p_{ch}}{p_{pump}}=\frac{7.78\times 10^{-4}}{8.33\times 10^{-5}}=9.33

about nine times the pump-side pressure.

Engineering Comment

This exercise explains a common false acceptance. A pump-side gauge can show a very good pressure while the chamber or test-article region is much higher because the connecting path has limited conductance. Acceptance pressure should be measured or inferred at the region that matters, not only at the easiest place to pump.

Plausibility Check

The chamber pressure must be higher than the pump-side pressure because gas must flow through a finite conductance. A factor of about nine is credible because the tube conductance is much smaller than the pump speed.

Exercise 8: Radiative Heat Rejection in Vacuum

An instrument stage in vacuum dissipates:

Q_{gen}=18\ \text{W}

It has a radiating area:

A=0.080\ \text{m}^2

with emissivity:

\epsilon=0.75

The stage temperature is:

T_h=320\ \text{K}

and chamber wall temperature is:

T_c=295\ \text{K}

Estimate radiative heat rejection. Determine how much heat must still be removed by conduction straps, mounts or other paths.

Solution

For a simplified gray-surface radiation screen:

Q_{rad}=\epsilon\sigma A(T_h^4-T_c^4)

Substitute:

Q_{rad}=0.75(5.670374419\times 10^{-8})(0.080)(320^4-295^4)
Q_{rad}=9.91\ \text{W}

Remaining heat to remove by other paths is:

Q_{remain}=Q_{gen}-Q_{rad}
Q_{remain}=18-9.91=8.09\ \text{W}

Engineering Comment

In high vacuum, gas convection is usually not available as a design assumption. This stage cannot rely on radiation alone at the stated temperatures and area. The design needs a conduction path through straps, mounts, harnesses or controlled contact interfaces, and the validation should measure stage temperature under vacuum, not only in air.

Plausibility Check

Radiation removes a little more than half of the heat load. Since the temperature difference is only 25\ \text{K} and the area is modest, it would be suspicious if the radiation calculation alone removed the full 18\ \text{W}.

Exercise 9: Pumpdown Regression After Venting

A vacuum chamber normally reaches:

p_{base}=1.0\times 10^{-3}\ \text{Pa}

after:

t_{base}=6.0\ \text{min}

following a standard dry-nitrogen vent and pumpdown. After maintenance, the chamber pressure at the same elapsed time is:

p_{6}=2.8\times 10^{-3}\ \text{Pa}

and the chamber does not reach 1.0\times 10^{-3}\ \text{Pa} until:

t_{new}=14.5\ \text{min}

The acceptance rule allows a pumpdown-time ratio no greater than:

R_{max}=1.5

or, as an alternate screen, pressure at the baseline time no greater than:

p_{6,max}=1.5\times 10^{-3}\ \text{Pa}

Evaluate the regression.

Solution

The time-ratio regression is:

\displaystyle R_t=\frac{t_{new}}{t_{base}}
\displaystyle R_t=\frac{14.5}{6.0}=2.42

Compare with the limit:

2.42>1.5

The pumpdown-time screen fails.

The pressure excess at the baseline time is:

\displaystyle R_p=\frac{p_6}{p_{base}}
\displaystyle R_p=\frac{2.8\times 10^{-3}}{1.0\times 10^{-3}}=2.8

The alternate pressure criterion also fails because:

2.8\times 10^{-3}>1.5\times 10^{-3}\ \text{Pa}

The chamber should not be released without investigating the maintenance effect.

Engineering Comment

This is a practical recovery check after venting, cleaning, fixture change or seal replacement. The absolute pressure eventually reaches the old target, but the recovery curve has changed enough to indicate added gas load, trapped volume, water adsorption, partially closed conductance path, gauge contamination or pump performance loss. A release decision should compare the full curve, not only the final endpoint.

Plausibility Check

Both independent screens tell the same story: the chamber is more than twice as slow and almost three times too high at the baseline time. The result is not a borderline failure caused by rounding.

Exercise 10: Virtual Leak Volume from a Pressure Burst

An isolated chamber has volume:

V_{ch}=0.50\ \text{m}^3

and pressure:

p_{ch}=1.0\times 10^{-3}\ \text{Pa}

A trapped pocket with volume:

V_t=25\ \text{cm}^3

is initially at:

p_t=100\ \text{Pa}

and then opens suddenly into the chamber. Estimate the final pressure assuming isothermal mixing and no pumping during the burst. Interpret whether this is consistent with a finite virtual leak rather than a continuous external leak.

Solution

Convert trapped volume:

V_t=25\times 10^{-6}\ \text{m}^3

For isothermal mixing, total gas inventory gives:

\displaystyle p_f=\frac{p_{ch}V_{ch}+p_tV_t}{V_{ch}+V_t}

Substitute:

\displaystyle p_f=\frac{(1.0\times 10^{-3})(0.50)+(100)(25\times 10^{-6})}{0.50+25\times 10^{-6}}

The numerator is:

p_{ch}V_{ch}+p_tV_t=5.0\times 10^{-4}+2.5\times 10^{-3}=3.0\times 10^{-3}\ \text{Pa m}^3

Therefore:

\displaystyle p_f=\frac{3.0\times 10^{-3}}{0.500025}=6.00\times 10^{-3}\ \text{Pa}

The pressure burst is:

\Delta p=p_f-p_{ch}
\Delta p=6.00\times 10^{-3}-1.0\times 10^{-3}=5.00\times 10^{-3}\ \text{Pa}

Engineering Comment

A small trapped volume at modest pressure can create a visible high-vacuum pressure burst. If the pressure spike decays after pumping and does not repeat with the same slope, it is more consistent with a finite virtual leak or trapped volume than with a continuous external leak. The fix is often vent-pathing, drilled relief holes, cleaned threaded interfaces or fixture redesign, not simply more pump speed.

Plausibility Check

The trapped pocket volume is only 0.005\% of the chamber volume, but its pressure is 100000 times higher than the chamber pressure. A millipascal-scale jump is therefore plausible and should not be dismissed because the pocket is physically small.

Exercise 11: Gas-Species Gauge Correction

An ionization gauge is calibrated for nitrogen and reads:

p_{ind}=3.0\times 10^{-4}\ \text{Pa}

The process limit is:

p_{limit}=8.0\times 10^{-4}\ \text{Pa}

Use relative sensitivity factors:

S_{Ar}=1.3

for argon and:

S_{He}=0.18

for helium, where:

\displaystyle p_{true}=\frac{p_{ind}}{S_g}

Evaluate whether the same indicated pressure passes the process limit if the dominant gas is nitrogen, argon or helium.

Solution

For nitrogen calibration gas:

S_{N2}=1.0
\displaystyle p_{N2}=\frac{3.0\times 10^{-4}}{1.0}=3.0\times 10^{-4}\ \text{Pa}

For argon:

\displaystyle p_{Ar}=\frac{3.0\times 10^{-4}}{1.3}=2.31\times 10^{-4}\ \text{Pa}

For helium:

\displaystyle p_{He}=\frac{3.0\times 10^{-4}}{0.18}=1.67\times 10^{-3}\ \text{Pa}

Compare with the limit:

p_{N2}<p_{limit}
p_{Ar}<p_{limit}
p_{He}>p_{limit}

The nitrogen and argon interpretations pass the stated pressure limit, while the helium-dominated interpretation fails.

Engineering Comment

Species correction can change the release decision. A nitrogen-calibrated ion gauge reading is not an absolute truth when the gas mixture changes after helium leak testing, purge gas exposure, process gas operation or desorption. Acceptance evidence should state the assumed dominant gas or use residual-gas analysis when species composition matters.

Plausibility Check

The helium correction increases pressure because the gauge is much less sensitive to helium than to nitrogen. A corrected helium pressure more than five times the indicated reading is plausible with the stated sensitivity factor.

Exercise 12: Guarded Local Pressure Release

A local gauge near the test article measures:

p_m=7.2\times 10^{-4}\ \text{Pa}

The release limit is:

p_{max}=1.0\times 10^{-3}\ \text{Pa}

Estimate a relative uncertainty budget with standard components:

SourceStandard relative uncertainty
Gauge calibration8\%
Repeatability5\%
Conductance inference10\%
Temperature and gas correction6\%

Use root-sum-square combination and expanded coverage factor:

k=2

Calculate the guarded upper pressure and decide whether the release passes.

Solution

Combine relative standard uncertainty:

u_r=\sqrt{0.08^2+0.05^2+0.10^2+0.06^2}
u_r=\sqrt{0.0064+0.0025+0.0100+0.0036}=0.150

Expanded relative uncertainty is:

U_r=ku_r
U_r=2(0.150)=0.300

Expanded absolute uncertainty is:

U_p=U_rp_m
U_p=0.300(7.2\times 10^{-4})=2.16\times 10^{-4}\ \text{Pa}

The guarded upper pressure is:

p_{guard}=p_m+U_p
p_{guard}=7.2\times 10^{-4}+2.16\times 10^{-4}=9.36\times 10^{-4}\ \text{Pa}

Compare with the limit:

9.36\times 10^{-4}<1.0\times 10^{-3}\ \text{Pa}

The guarded release passes, but the margin is:

M=p_{max}-p_{guard}
M=1.0\times 10^{-3}-9.36\times 10^{-4}=6.4\times 10^{-5}\ \text{Pa}

Engineering Comment

The measured pressure alone appears comfortably below the limit, but the guarded result is close. This is the correct release logic for a local test-article requirement: include gauge calibration, repeatability, conductance inference and gas correction before declaring margin. If the next configuration changes gas species, temperature or gauge placement, this release evidence should not be reused without review.

Plausibility Check

The combined standard uncertainty is 15\%, and the expanded uncertainty is 30\%, so a guarded upper pressure about 30\% above the measured value is expected. The pass is real under the stated model, but the margin is only 6.4\% of the limit.

Exercise 13: Turbopump Crossover Pressure and Foreline Interlock

A chamber is rough pumped before a turbomolecular pump is allowed to start. The chamber volume is:

V=0.80\ \text{m}^3

Initial pressure after venting is:

p_0=1.0\times10^5\ \text{Pa}

The roughing path has effective pumping speed at the chamber:

S_r=0.012\ \text{m}^3/\text{s}

The turbopump start rule requires chamber pressure below:

p_x=8.0\ \text{Pa}

The backing pump speed at the foreline is:

S_b=0.0040\ \text{m}^3/\text{s}

The foreline interlock limit is:

p_{fore,max}=50\ \text{Pa}

Use the ideal roughing estimate:

\displaystyle t=\frac{V}{S_r}\ln\left(\frac{p_0}{p_x}\right)

and approximate gas throughput near crossover as:

Q=pS_r

with foreline pressure:

\displaystyle p_{fore}=\frac{Q}{S_b}

Estimate the ideal roughing time to the crossover pressure. Then check foreline pressure if the turbopump is started at 8.0\ \text{Pa} and if an operator tries to start early at 20\ \text{Pa}.

Solution

Ideal roughing time to 8.0\ \text{Pa}:

\displaystyle t=\frac{0.80}{0.012}\ln\left(\frac{1.0\times10^5}{8.0}\right)
t=66.7\ln(12500)
t=629\ \text{s}

Convert to minutes:

\displaystyle t=\frac{629}{60}=10.5\ \text{min}

Throughput at the correct crossover pressure:

Q_x=p_xS_r=8.0(0.012)=0.096\ \text{Pa m}^3/\text{s}

Foreline pressure at crossover:

\displaystyle p_{fore,x}=\frac{0.096}{0.0040}=24\ \text{Pa}

Since:

24<50\ \text{Pa}

the foreline interlock passes at the specified crossover pressure.

If the operator starts early at 20\ \text{Pa}:

Q_{20}=20(0.012)=0.240\ \text{Pa m}^3/\text{s}

Foreline pressure becomes:

\displaystyle p_{fore,20}=\frac{0.240}{0.0040}=60\ \text{Pa}

Since:

60>50\ \text{Pa}

the early start fails the foreline interlock.

Engineering Comment

Turbopump crossover is not only a chamber-pressure threshold. The backing path must also handle the gas throughput without exceeding foreline pressure, overheating, backstreaming or driving the turbopump outside its safe start envelope. A real procedure should validate gauge location, roughing valve state, backing-pump capacity, vent history, water vapor load, turbo controller interlocks and abort logic during failed-start tests.

Plausibility Check

The ideal roughing time is about ten minutes because the pressure ratio from atmosphere to 8\ \text{Pa} is large. Starting at 20\ \text{Pa} instead of 8\ \text{Pa} multiplies throughput by 20/8=2.5, so the foreline pressure rises from 24\ \text{Pa} to 60\ \text{Pa}. That proportional change confirms the interlock failure is driven by gas throughput, not by arithmetic noise.

Exercise 14: Residual-Gas Partial-Pressure Acceptance

A process chamber reaches total pressure:

p_{tot}=8.5\times10^{-4}\ \text{Pa}

after pumpdown. Residual-gas analysis estimates that water vapor is:

x_{H2O}=0.42

of the total pressure, and hydrocarbon fragments are:

x_{HC}=0.08

of the total pressure. The process release limits are:

p_{H2O,max}=3.0\times10^{-4}\ \text{Pa}

and:

p_{HC,max}=5.0\times10^{-5}\ \text{Pa}

Calculate water and hydrocarbon partial pressures and decide whether the chamber passes. Then repeat the decision after bakeout gives:

p_{tot,b}=7.2\times10^{-4}\ \text{Pa},\quad x_{H2O,b}=0.22,\quad x_{HC,b}=0.04

Solution

Water partial pressure before bakeout:

p_{H2O}=x_{H2O}p_{tot}
p_{H2O}=0.42(8.5\times10^{-4})=3.57\times10^{-4}\ \text{Pa}

Hydrocarbon partial pressure before bakeout:

p_{HC}=0.08(8.5\times10^{-4})=6.8\times10^{-5}\ \text{Pa}

Both limits fail:

3.57\times10^{-4}>3.0\times10^{-4}

and:

6.8\times10^{-5}>5.0\times10^{-5}

After bakeout:

p_{H2O,b}=0.22(7.2\times10^{-4})=1.58\times10^{-4}\ \text{Pa}
p_{HC,b}=0.04(7.2\times10^{-4})=2.88\times10^{-5}\ \text{Pa}

Both partial-pressure limits pass after bakeout.

Engineering Comment

Total pressure can pass while the gas composition still fails the process requirement. Water vapor may control surface chemistry, optics, insulation, plasma ignition or contamination even when the chamber pressure looks acceptable. Hydrocarbon fragments may indicate pump oil, elastomers, cleaning residue or backstreaming risk. A release package should state RGA calibration, cracking-pattern interpretation, background subtraction, gauge location and whether the RGA represents the process region.

Plausibility Check

The initial total pressure is below 1.0\times10^{-3}\ \text{Pa}, but 42\% water makes the water partial pressure 3.57\times10^{-4}\ \text{Pa}, just above the limit. Bakeout lowers both total pressure and contaminant fractions, so the water partial pressure falls by more than half and the hydrocarbon partial pressure also passes. That direction is physically plausible.

Exercise 15: Nitrogen Purge Dilution Before Pumpdown

A chamber with volume:

V=0.60\ \text{m}^3

is purged with dry nitrogen before pumpdown. Treat the chamber as well mixed. The purge flow at chamber conditions is:

Q=0.030\ \text{m}^3/\text{min}

Initial oxygen concentration is:

C_0=21\%

The process requirement before pumpdown is:

C_{target}=1.0\%

Use the well-mixed dilution model:

\displaystyle C=C_0\exp\left(-\frac{Qt}{V}\right)

Calculate purge time to reach the target, equivalent volume exchanges and nitrogen volume used. Then calculate oxygen concentration after only 30\ \text{min}.

Solution

Solve for purge time:

\displaystyle t=\frac{V}{Q}\ln\left(\frac{C_0}{C_{target}}\right)
\displaystyle t=\frac{0.60}{0.030}\ln\left(\frac{21}{1.0}\right)
t=20(3.045)=60.9\ \text{min}

Equivalent well-mixed volume exchanges:

\displaystyle N_{ex}=\frac{Qt}{V}=\frac{0.030(60.9)}{0.60}=3.05

Nitrogen volume used:

V_{N2}=Qt=0.030(60.9)=1.83\ \text{m}^3

After only 30\ \text{min}:

\displaystyle C_{30}=21\exp\left(-\frac{0.030(30)}{0.60}\right)
C_{30}=21\exp(-1.5)=4.69\%

The 30\ \text{min} purge does not meet the 1.0\% oxygen target.

Engineering Comment

A nitrogen purge is not automatically equivalent to a clean inert state. Flow path, dead zones, fixture cavities, purge port placement, exhaust restriction, humidity, oxygen analyzer location and mixing behavior can all change the real dilution curve. If the target protects a sensitive process, the purge should be validated with oxygen or moisture measurements at the relevant location, not only with elapsed time.

Plausibility Check

Three well-mixed volume exchanges reduce concentration by about \exp(-3)=0.05, so reducing oxygen from 21\% to about 1\% is plausible. A 30\ \text{min} purge gives only 1.5 volume exchanges and leaves about 4.7\% oxygen, so treating half the time as sufficient would be unsafe.

Exercise 16: Pump-Trip Hold Time and Guarded Abort Decision

A vacuum process must remain below a local pressure limit after a pump trip long enough for the tool to close shutters and move the test article to a safe state. The protected volume is:

V=0.25\ \text{m}^3

The pressure at the start of the pump-trip event is:

p_0=8.0\times10^{-4}\ \text{Pa}

The process pressure limit is:

p_{max}=5.0\times10^{-3}\ \text{Pa}

The safe-state sequence requires:

t_{req}=12\ \text{min}

The measured gas-load contributors are:

Q_{out}=6.0\times10^{-7}\ \text{Pa m}^3/\text{s}
Q_{leak}=1.2\times10^{-7}\ \text{Pa m}^3/\text{s}
Q_{valve}=4.0\times10^{-7}\ \text{Pa m}^3/\text{s}

Assume the chamber is isolated and well mixed during the event, so pressure rise follows:

\displaystyle \frac{dp}{dt}=\frac{Q}{V}

Compute nominal hold time to the pressure limit. Then apply a release guard using 25\% higher gas load and a pressure-limit guard of:

p_{guard}=4.7\times10^{-3}\ \text{Pa}

Finally, evaluate a corrective action that reduces valve seepage to:

Q_{valve,new}=1.0\times10^{-7}\ \text{Pa m}^3/\text{s}

Solution

Nominal total gas load:

Q=Q_{out}+Q_{leak}+Q_{valve}
Q=6.0\times10^{-7}+1.2\times10^{-7}+4.0\times10^{-7}=1.12\times10^{-6}\ \text{Pa m}^3/\text{s}

Nominal pressure rise allowed before reaching the process limit:

\Delta p=p_{max}-p_0
\Delta p=5.0\times10^{-3}-8.0\times10^{-4}=4.2\times10^{-3}\ \text{Pa}

Nominal hold time:

\displaystyle t_{hold}=\frac{V\Delta p}{Q}
\displaystyle t_{hold}=\frac{0.25(4.2\times10^{-3})}{1.12\times10^{-6}}=938\ \text{s}

Convert to minutes:

\displaystyle t_{hold}=\frac{938}{60}=15.6\ \text{min}

Nominal margin above the required safe-state sequence is:

M_t=15.6-12=3.6\ \text{min}

The nominal calculation passes.

For the guarded release, use:

Q_g=1.25Q
Q_g=1.25(1.12\times10^{-6})=1.40\times10^{-6}\ \text{Pa m}^3/\text{s}

Guarded pressure rise:

\Delta p_g=p_{guard}-p_0
\Delta p_g=4.7\times10^{-3}-8.0\times10^{-4}=3.9\times10^{-3}\ \text{Pa}

Guarded hold time:

\displaystyle t_{hold,g}=\frac{0.25(3.9\times10^{-3})}{1.40\times10^{-6}}=696\ \text{s}
t_{hold,g}=11.6\ \text{min}

The guarded result fails the required 12\ \text{min} sequence by:

12-11.6=0.4\ \text{min}

Now reduce valve seepage:

Q_{new}=6.0\times10^{-7}+1.2\times10^{-7}+1.0\times10^{-7}=8.2\times10^{-7}\ \text{Pa m}^3/\text{s}

Guarded new gas load:

Q_{g,new}=1.25(8.2\times10^{-7})=1.03\times10^{-6}\ \text{Pa m}^3/\text{s}

Corrected guarded hold time:

\displaystyle t_{hold,new}=\frac{0.25(3.9\times10^{-3})}{1.03\times10^{-6}}=951\ \text{s}
t_{hold,new}=15.9\ \text{min}

The corrected valve-seepage case passes the 12\ \text{min} guarded requirement with about 3.9\ \text{min} margin.

Engineering Comment

Pump-trip acceptance is a time-to-failure problem, not just a steady pressure problem. The nominal pressure may look safe while the guarded hold time is too short for the abort sequence. Release evidence should include the measured gas-load split, isolation valve state, pressure sensor location and response time, controller latency, pressure-limit guard, safe-state timing and a repeated pump-trip test if the consequence is high.

Plausibility Check

The pressure window is only a few millipascals, so a gas load near 10^{-6}\ \text{Pa m}^3/\text{s} in a 0.25\ \text{m}^3 volume should give a hold time of order 10^3\ \text{s}. The guarded result is shorter because the allowed pressure rise is smaller and the gas load is larger. Reducing the largest controllable contributor, valve seepage, restores several minutes of margin.

Exercise 17: Thermal Transpiration Pressure Correction for a Hot Volume

A thermal-vacuum fixture has a room-temperature ion gauge near the chamber wall. The gauge temperature is:

T_g=295\ \text{K}

and the indicated nitrogen-corrected pressure is:

p_g=6.0\times10^{-4}\ \text{Pa}

The accepted volume is a small test-article cavity operating in molecular flow at:

T_h=520\ \text{K}

The local hot-cavity pressure limit is:

p_{lim}=8.0\times10^{-4}\ \text{Pa}

Use the molecular-flow thermal-transpiration screen:

\displaystyle p_h=p_g\sqrt{\frac{T_h}{T_g}}

where p_h is the pressure estimate for the hot accepted volume. Compute the corrected hot-cavity pressure and the nominal margin. Then apply a guarded release pressure:

p_{rel}=1.15p_h

and decide whether the room-temperature gauge reading supports release. Finally, repeat the guarded decision if a bakeout and purge step lowers the room-temperature gauge reading to:

p_{g,new}=5.0\times10^{-4}\ \text{Pa}

Solution

Thermal-transpiration factor:

\displaystyle F_T=\sqrt{\frac{T_h}{T_g}}
\displaystyle F_T=\sqrt{\frac{520}{295}}=1.33

Corrected hot-cavity pressure:

p_h=6.0\times10^{-4}(1.33)=7.97\times10^{-4}\ \text{Pa}

Nominal margin to the hot-cavity limit:

M_p=p_{lim}-p_h
M_p=8.0\times10^{-4}-7.97\times10^{-4}=3.4\times10^{-6}\ \text{Pa}

The nominal result barely passes. The room-temperature gauge reading alone looks more comfortable:

8.0\times10^{-4}-6.0\times10^{-4}=2.0\times10^{-4}\ \text{Pa}

but that margin does not represent the hot local volume.

Guarded release pressure:

p_{rel}=1.15(7.97\times10^{-4})=9.16\times10^{-4}\ \text{Pa}

Since:

9.16\times10^{-4}>8.0\times10^{-4}

the hot-volume acceptance fails after the guard is applied.

For the improved gauge reading:

p_{h,new}=5.0\times10^{-4}(1.33)=6.64\times10^{-4}\ \text{Pa}

Guarded improved pressure:

p_{rel,new}=1.15(6.64\times10^{-4})=7.63\times10^{-4}\ \text{Pa}

Now:

7.63\times10^{-4}<8.0\times10^{-4}

so the improved case passes this guarded hot-volume pressure screen.

Engineering Comment

Thermal transpiration matters when the pressure decision belongs to a hot molecular-flow region but the gauge is at a colder boundary. The calculation is still a screen: the release record should state gas species, Knudsen regime, gauge correction, aperture geometry, temperature stability, gas-load source and whether the hot cavity has reached equilibrium. A total-pressure pass at a cold gauge is not enough evidence for a hot local acceptance limit.

Plausibility Check

The hot volume is at 520\ \text{K}, much warmer than the 295\ \text{K} gauge, so the corrected local pressure should be higher than the gauge reading. The factor is only about 1.33, not an order of magnitude, so the correction is plausible but still large enough to erase a small acceptance margin.

Exercise 18: Water-Vapor Exposure Dose and Surface Release

A surface-sensitive vacuum process has a water-vapor exposure limit before processing. Residual-gas analysis estimates water partial pressure near the fixture as:

p_{H2O}=4.0\times10^{-6}\ \text{Pa}

The surface waits at this condition for:

t=45\ \text{min}

Use the screening convention:

1\ \text{Langmuir}=1.0\times10^{-6}\ \text{Torr s}

The guarded release rule is:

E_{H2O}+G_E\leq E_{lim}

with:

E_{lim}=50\ \text{Langmuir}

and:

G_E=10\ \text{Langmuir}

Calculate water-vapor exposure and decide whether the process can be released. Then check an improved bake-and-purge condition with:

p_{H2O,new}=1.1\times10^{-6}\ \text{Pa}

and:

t_{new}=30\ \text{min}

Solution

Convert water partial pressure to Torr:

\displaystyle p_{H2O,Torr}=\frac{4.0\times10^{-6}}{133.322}=3.00\times10^{-8}\ \text{Torr}

Convert time:

t=45(60)=2700\ \text{s}

Exposure in Torr seconds:

p_{H2O,Torr}t=(3.00\times10^{-8})(2700)=8.10\times10^{-5}\ \text{Torr s}

Exposure in Langmuir:

\displaystyle E_{H2O}=\frac{8.10\times10^{-5}}{1.0\times10^{-6}}=81.0\ \text{Langmuir}

Guarded exposure:

E_{guard}=E_{H2O}+G_E=81.0+10=91.0\ \text{Langmuir}

The process should not be released because:

91.0>50

Now check the improved condition. Convert pressure:

\displaystyle p_{H2O,new,Torr}=\frac{1.1\times10^{-6}}{133.322}=8.25\times10^{-9}\ \text{Torr}

Convert time:

t_{new}=30(60)=1800\ \text{s}

Improved exposure:

\displaystyle E_{H2O,new}=\frac{(8.25\times10^{-9})(1800)}{1.0\times10^{-6}}=14.9\ \text{Langmuir}

Guarded improved exposure:

E_{guard,new}=14.9+10=24.9\ \text{Langmuir}

The improved bake-and-purge condition passes the 50 Langmuir guarded exposure limit.

Engineering Comment

Surface contamination is time integrated. A water partial pressure that appears small can still deliver too much exposure if the fixture waits at that state for long enough. Total pressure, pumpdown endpoint and instantaneous RGA fraction do not by themselves prove that a surface-sensitive process is clean enough.

The Langmuir convention is a first-pass exposure metric, not a universal monolayer guarantee. Real sticking probability, surface temperature, material, prior contamination, plasma cleaning, line-of-sight geometry and re-desorption can all change the surface dose. Release evidence should state RGA location, water calibration basis, exposure start and stop time, surface temperature, accepted surface boundary and whether the limit is empirical for the actual process.

Plausibility Check

A water partial pressure of about 3\times10^{-8}\ \text{Torr} for 2700 seconds gives roughly 3\times2700\times10^{-2} Langmuir, or about 81 Langmuir. That already exceeds the 50 Langmuir limit before adding the guard. The improved case reduces both pressure and time, so the exposure drops to about 15 Langmuir and remains below the guarded limit.

Review Table

CheckResultInterpretation
Pressure conversion2.0\times 10^{-5}\ \text{mbar}, 1.50\times 10^{-5}\ \text{Torr}Pressure must be absolute and unit convention must be explicit.
Gas inventory2.97\times 10^{17} moleculesVacuum is low density, not zero gas.
Effective speed0.0261\ \text{m}^3/\text{s}Conductance reduces chamber speed to about 13\% of pump speed.
Ideal pumpdown5.9\ \text{min}A clean ideal model can understate real pumpdown time.
Plateau gas load7.83\times 10^{-5}\ \text{Pa m}^3/\text{s}Continuing gas load controls final pressure.
Rate-of-rise load2.0\times 10^{-6}\ \text{Pa m}^3/\text{s}Passes the stated gas-load acceptance limit.
Outgassing pressure3.45\times 10^{-4}\ \text{Pa}Passes target if the surface condition is real.
Knudsen number13.3 to 266Free molecular assumptions dominate the local features.
Pump-side errorchamber pressure about 9.3 times pump-side pressureGauge placement can control acceptance validity.
Radiative heat rejection9.91\ \text{W}Additional conductive heat removal is required.
Pumpdown regressiontime ratio 2.42, pressure ratio 2.8Maintenance recovery fails both practical release screens.
Virtual leak burstp_f=6.00\times 10^{-3}\ \text{Pa}A small trapped volume can create a finite pressure burst.
Gauge gas correctionhelium-corrected pressure 1.67\times 10^{-3}\ \text{Pa}The same indicated gauge reading can fail for a different dominant gas.
Guarded pressure releasep_{guard}=9.36\times 10^{-4}\ \text{Pa}The local pressure passes only after explicit uncertainty accounting.
Turbopump crossover10.5\ \text{min} to 8.0\ \text{Pa}; early start gives 60\ \text{Pa} forelineCrossover must satisfy both chamber and foreline limits.
RGA partial pressurewater 3.57\times10^{-4}\ \text{Pa} and hydrocarbons 6.8\times10^{-5}\ \text{Pa} before bakeoutTotal pressure alone can hide contaminant failures.
Nitrogen purge dilution60.9\ \text{min} and 3.05 volume exchanges to reach 1.0\% oxygenElapsed purge time must be tied to dilution evidence.
Pump-trip hold timenominal 15.6\ \text{min}, guarded 11.6\ \text{min}, corrected guarded 15.9\ \text{min}Abort timing must pass with pressure and gas-load guards, not only nominal gas load.
Thermal transpiration correctionhot-volume pressure 7.97\times10^{-4}\ \text{Pa}; guarded 9.16\times10^{-4}\ \text{Pa}Room-temperature gauge reading fails guarded hot-volume acceptance.
Water-vapor surface exposure81.0 Langmuir, guarded 91.0 LangmuirTime at partial pressure can fail surface cleanliness even when pressure looks small.

Common Mistakes

  • mixing absolute vacuum pressure with gauge pressure or pressure below atmosphere;
  • accepting pump nameplate speed instead of chamber effective pumping speed;
  • applying ideal pumpdown time after outgassing and virtual leaks dominate;
  • using a pump-side gauge as evidence for a local test-article environment;
  • treating a passing helium leak number as proof that outgassing is acceptable;
  • applying continuum flow or convection models when Knudsen number is large;
  • treating a finite virtual-leak burst as the same evidence as a continuous leak;
  • ignoring gas-species correction after helium testing, purge changes or process gas exposure;
  • accepting total pressure without checking residual-gas partial pressures that own contamination limits;
  • accepting a low water partial pressure without integrating exposure time for surface-sensitive processes;
  • treating nitrogen purge as complete without calculating volume exchanges or measuring oxygen/moisture;
  • accepting nominal pump-trip hold time without pressure-limit and gas-load guards;
  • accepting a room-temperature gauge reading for a hot molecular-flow volume without thermal-transpiration correction and temperature evidence;
  • using an unguarded gauge reading when the acceptance margin is small;
  • ignoring surface history, water vapor, elastomers, adhesives and warm fixtures;
  • validating pressure without recording valve state, gauge location, gas correction and temperature.

Review Checklist

Before accepting a vacuum calculation, confirm:

  1. pressure is absolute and tied to a stated gauge, range, calibration and gas correction;
  2. the accepted pressure boundary is the region that matters for the test article or process;
  3. pumpdown evidence includes valve state, conductance path and local pressure where needed;
  4. gas load is separated into external leak, outgassing, virtual leak and process gas where possible;
  5. rate-of-rise tests use stabilized temperature and documented isolation conditions;
  6. mean free path and Knudsen number are checked for the relevant local geometry;
  7. pumpdown regression is checked after venting, maintenance, material change or contamination event;
  8. gas-species correction is applied when the dominant gas may differ from the gauge calibration gas;
  9. guarded pressure release includes calibration, repeatability, conductance and gas-correction uncertainty;
  10. thermal calculations account for reduced gas cooling, radiation, conduction paths and measured temperatures;
  11. turbopump crossover procedures check chamber pressure, foreline pressure, roughing valve state and abort interlocks;
  12. residual-gas acceptance checks partial pressures for water, hydrocarbons, process gases and contaminants that own release limits;
  13. surface-sensitive release checks time-integrated exposure when water or contaminant partial pressure can accumulate dose;
  14. purge procedures state flow path, volume exchanges, oxygen or moisture target, analyzer location and acceptance evidence;
  15. pump-trip or isolation hold-time decisions include gas-load guard, pressure guard, controller latency and safe-state timing;
  16. hot local volumes use temperature-state or thermal-transpiration corrections when molecular flow makes a room-temperature gauge non-representative.
REF

See also